Download Sect 2.5 – Equations of Lines and Linear Models

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Sect 2.5 – Equations of Lines and Linear Models
Objective 1:
Point-Slope Formula
From the last section, we worked with the slope-intercept form of the linear
function:
Slope-intercept form
A linear equation in two variables is said to be written in slope-intercept
form if it is in the form y = mx + b where m is the slope of the line and the
point (0, b) is the y-intercept.
In this section, we want to introduce the point-slope form of a linear
function. To see where the point-slope form comes from, let us examine the
formula for calculating the slope:
m=
€
€
€
y 2 − y1
x2 − x1
y 2 − y1
=m
x2 − x1
y − y1
=m
x− x1
y − y1 (x− x1)
x− x1
1
(rewrite the formula)
(replace (x2, y2) with the variables x and y)
(clear fractions by multiply both sides by (x – x1))
= m(x – x1)
(reduce)
y – y1 = m(x – x1) This is a point-slope formula.
Point-Slope Form
€
€ A linear
equation in two variables is said to be written in point-slope form
if it is in the form: y – y1 = m(x – x1), where m is the slope and (x1, y1)
is any known point on the line.
Given the following information, find and write the equation in
a) point-slope form, b) slope-intercept form, and c) standard form.
Ex. 1 A line with slope of 6
Ex. 2 A line with slope of –
2
3
passing through the
point (9, 5).
Solution:
m = 6; (x1, y1) = (9, 5)
passing through the
point (– 8, 4).
Solution:
€
2
m = – ; (x1, y1) = (– 8, 4)
y – y1 = m(x – x1)
y – y1 = m(x – x1)
3
€
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2
(x – (– 8)) pt-slope
3
2
y – 4 = – (x + 8)
3
2
16
y–4=– x–
3
3
€
12
+4=
+
3
€
2
4
b) y = – x –
slope-int
3
€
€3
2
4
3(y) = 3 – x – 3
3
3
€
a) y – 5 = 6(x – 9) pt-slope
a) y – 4 = –
y – 5 = 6x – 54
+5=
+5
b) y = 6x – 49
slope-int
y = 6x – 49
(
– 6x = – 6x
– 6x + y = – 49
–1
–1
c) 6x – y = 49
y 2 − y1
x2 − x1
2
2
=–
−7
7
m=
=
=
.
3y = – 2x – 4
€+ 2x =€ + 2x
c) 2x + 3y = – 4 std form
€
€
std form
Ex. 3 A line that passes
through the points
(2, – 3) and (– 5, – 1).
Solution:
First, find the slope:
Ex. 4 A line that passes
through the points
(7, – 10) and (5, – 14).
Solution:
First, find the slope:
−1 − (−3)
−5−2
€
a) y – (– 3) = –
€
y+3=–
2
7
2
7
−4
−2
y 2 − y1
x2 − x1
=
−14−(−10)
5−7
= 2.
Now, pick one of the points.
€
m = 2; (x1, y1) = (7, – 10)
y – y1 = m(x – x1)
(x – 2) pt-slope a) y – (– 10) = 2(x – 7) pt-slope
4
7
21
7
x+
–3=
–
€
2
17
b) y = – x –
slope-int
7
7
€
€
2
17
7y = 7 – x – 7
7
€ 7
7y = – 2x – 17
€+ 2x =
€ + 2x
c) 2x + 7y = – 17 std form
€
€
(
m=
=
Now, pick one of the points.
€ 2
€
m = – ; (x1, y1) = (2, – 3)
7
€
y€– y1 = m(x – x1)
€
) ( )
) ( )
y + 10 = 2x – 14
– 10 =
– 10
b) y = 2x – 24 slope-int
– 2x = – 2x
– 2x + y = – 24
–1
–1
c) 2x – y = 24 std form
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Find the slope and y-intercept and sketch the graph:
Ex. 5
8x + 3y = 24
Solution:
First solve the equation for y:
8x + 3y = 24
– 8x
= – 8x
3y = – 8x + 24
3
3
8
y=– x+8
The slope is –
8
3
3
=
−8
3
=
"rise"
"run"
and the y-intercept is (0, 8).
Plot the €
point (0, 8). Then from
that point fall 8 units and run
€ another
€
3 units€to get
point.
From that new point fall another 8 units and run 3 more units to get
the third point. Now, draw the graph.
For each graph, write the equation of the line:
Ex. 6
Ex. 7
Cost
of a
TV
Cost
of a
gallon
of gas
Year
Month
Solution:
Two points on the graph are
(2, 1000) and (6, 500). The
500−1000
slope is then
= – 125.
Solution:
Two points on the graph are
(0, 3) and (5, 4). The slope is
4−3
then
= 0.2. The point
The point (0, 1250) is the
y-intercept. So, m = – 125 and
b = 1250. Thus, the equation of
€ is y = – 125x + 1250. As
the line
a linear function, we can write
f(x) = – 125x + 1250.
(0, 3) is the y-intercept. So,
m = 0.2 and b = 3. Thus, the
the equation of the line is
y = 0.2x + 3. As a linear
function, we can write
f(x) = 0.2x + 3.
6−2
5−0
€
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As mentioned before, the two special cases for linear equations were
vertical and horizontal lines.
Equations of Vertical and Horizontal lines
1) An equation of a vertical line passing through the point (a, b) is x = a.
The slope of such line is undefined.
2) An equation of a horizontal line passing through the point (a, b) is y = b.
The slope of such line is 0.
Use the given information to write the equation of the line:
Ex. 8
The slope of the line is 0 and it passes through (– 2, 5).
Solution:
m = 0, but we do not have a y-intercept. A line with a slope of 0 is a
horizontal line. Recall that a horizontal is in the form of y = constant.
Since it has to pass through the point (– 2, 5), then y has to be 5.
So, the equation is y = 5.
Ex. 9
The slope of the line is undefined and it passes through
(– 2, 5).
Solution:
m is undefined and we do not have a y-intercept. A line with a slope
that is undefined is a vertical line. Recall that a vertical is in the form
of x = constant. Since it has to pass through the point (– 2, 5), then
x has to be – 2.
So, the equation is x = – 2.
Objective 2:
Parallel and Perpendicular Lines.
Two distinct lines are parallel if and only if they have the same slope. We
can denote the slope of a parallel line by m||.
Two distinct lines that are neither vertical nor horizontal lines are
perpendicular if and only if the product of their slopes is – 1. In other
words, the slope of one line is the negative reciprocal of the slope of the
line perpendicular to it so long as they are not vertical or horizontal lines.
Every vertical line is perpendicular to every horizontal line and every
horizontal line is perpendicular to every vertical line. We can denote the
slope of a perpendicular line by m⊥.
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Write the equation of the line in slope-intercept form:
Ex. 10
Find the equation of the line passing through (– 2, 6).
a) parallel to y = 3x – 2.
b) perpendicular to y = 3x – 2.
Solution:
a) Parallel lines have the same slope, so m|| = 3.
y – y1 = m(x – x1)
y – 6 = 3(x – (– 2)) = 3(x + 2) = 3x + 6
y – 6 = 3x + 6
+6= +6
y = 3x + 12
b) The slope of the line perpendicular is the negative reciprocal of
1
3. So, m⊥ = – .
3
y – y1 = m(x – x1)
1
y – 6 = – (x – (– 2)) = –
y – 6€= –
3
1
x
3
–
+6=
+
€
1
16
y=– x+
3
3
€
€
1
(x
3
2
3
18
3€
+ 2) = –
€
1
x
3
–
2
3
€
€ equation of the line passing through (4, 0)
Ex. 11
Find the
1
a) €parallel
to
y
=
x – 3.
€
6
b) perpendicular to y =
1
x
6
– 3.
Solution:
€ lines have the same slope, so m|| =
a) Parallel
€ – x1 )
y – y1 = m(x
1
y – 0 = (x – 4) =
6
y=
1
x
6
–
1
x
6
–
4
6
=
1
x
6
2
3
–
1
.
6
2
3
€
b) The slope of the line perpendicular is the negative reciprocal of
€1 . So, m €
€ €
€
⊥ = – 6.
6
€
€
y€– y1 = m(x – x1)
y – 0 = – 6(x – 4) = – 6x + 24
y = – 6x + 24
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Ex. 12
Find the equation of the line passing through (8, – 3)
a) parallel to x = 5.
b) perpendicular to x = 5.
Solution:
a) A line parallel to x = 5 is a vertical line and it passes through
(8, – 3). So, the equation is x = 8.
b) A line perpendicular to x = 5 is a horizontal line and it passes
through (8, – 3). So, the equation is y = – 3.
Objective 3:
Modeling Data that Changes at a Constant Rate.
If the data in an application problem changes at nearly a constant rate, we
can use a linear function to approximate the data and this gives us a model
for the data where the average rate of change is the slope of the line. We
will typically draw what is called a trend line to model the data and pick two
points that lie the closest on the line to derive the equation of the model.
Find the equation of the line.
Ex. 13
The Average annual cost of tuition, board, fees, etc at a public
four-year institution is given in the graph below where x represents the
number of years after 2001 and y represents the cost.
$17,860
Cost
$12,300
Source: www.collegeboard.org
Years after 2001
a)
b)
Find a linear function that models the data
Use the model found in part a to predict the cost in 2016.
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Solution:
a)
The two points given on the graph are (1, 12300) and
(11, 17860). The slope is
y 2 − y1
x2 − x1
=
17860−12300
11−1
=
5560
10
= 556.
This means that the cost is increasing at an average rate of
$556 a year. To find the equation of the line, we can use the
point-slope formula:
€
€– x ) €
y – y1 = m(x
1
y – 12300 = 556(x – 1)
y – 12300 = 556x – 556
+ 12300 = + 12300
y = 556x + 11744
Thus, the linear function that models the data is
f(x) = 556x + 11,744
b)
Since x represents the number of years after 2001, then x = 1
represents 2002, x = 2 represents 2003, etc. Thus, x = 15
represents 2016. Evaluating f at x = 15 yields:
f(15) = 556(15) + 11,744 = 8340 + 11,744 = 20,084
The model predicts that the cost will be ≈ $20,084 in the year
2016.
It is worth noting that if we had use different data points, we would have
derived a slightly different answer.
Below is a summary of different formulas we have seen involving linear
equations.
Form
Standard Form
Ax + By = C
Horizontal Line
y = constant
Vertical Line
x = constant
Slope-Intercept
(Function) Form
f(x) = y = mx + b
Point-Slope Form
y – y1 = m(x – x1)
Example
5x – 3y = 17
y = – 2.3
x=3
m = 3, (0, b) = (0, – 5)
f(x) = y = 3x – 5
m = – 4, (x1, y1) = (2, 3)
y – 3 = – 4(x – 2)
Notes
A and B are integers &
must not both be 0.
m = 0,
y-int: (0, constant)
m is undefined
This is not a function.
m is the slope,
(0, b) is the y-int.
m is the slope, (x1, y1)
is any point on the line.